Derivative of (x^2+x)^(1/2): Simple Solution Manual

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D= (x^2+x)^(1/2)

I have a solutions manual here showing
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(x')
but correct me if I am wrong..
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(2x')
 
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well you didn't state what you were taking derivatives with respect to.

But if you have D(t)=f(x)

then D'(t)=f'(x)* (dx/dt) = f'(x)*x'

In your question; f(x)=\sqrt{x^2+x}
 
differentiating with respect to time t if you could please answer this simple question directly i would be glad especialy
 
nameVoid said:
D= (x^2+x)^(1/2)

I have a solutions manual here showing
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(x')
but correct me if I am wrong..
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(2x')

The first one is correct; the second one is incorrect.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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